Mean from a Frequency Table

A survey asks 30 households how many cars they own. Nobody wants to write the number 2 down twelve separate times, so the results arrive as a frequency table: "5 households have 1 car, 12 have 2 cars, 13 have 3 cars".

Now find the mean number of cars. Tempting shortcut: average the values 1, 2, 3 to get 2 — but that's wrong. It pretends one household owns each amount, when really 13 of them own three cars. The many three-car homes should pull the mean upward. The fix is to weight each value by how often it happens.

The method: build an f \times x column

When data comes as a frequency table, each value is listed once with how often it occurs. The mean is still "total ÷ how many", but a value that occurs ten times must be counted ten times over. So multiply each value by its frequency:

\text{mean} = \frac{\sum (\text{value} \times \text{frequency})}{\sum \text{frequency}} = \frac{\sum f x}{\sum f}

Add an f \times x column, total it to get \sum f x, total the frequencies to get \sum f, and divide. Here is a small worked table where value 1 occurs 4 times, 2 occurs 6 times and 3 occurs 10 times:

Value (x)	Frequency (f)	f \times x
1	4	4
2	6	12
3	10	30
Total	20	46

Here \sum f = 4 + 6 + 10 = 20 (20 items in all) and \sum f x = (1\times 4) + (2\times 6) + (3\times 10) = 46. So the mean is 46 \div 20 = 2.3.

Worked example 1 — goals per match

A football team's goals in their last 20 matches:

Goals (x)	Matches (f)	f \times x
0	6	0
1	7	7
2	4	8
3	3	9
Total	20	24

\sum f = 6 + 7 + 4 + 3 = 20 matches, and \sum f x = 0 + 7 + 8 + 9 = 24 goals. The mean is 24 \div 20 = 1.2 goals per match.

Notice the row 0 \times 6 = 0 still counts: those six no-score matches drag the mean down, which is exactly right. And 1.2 is not a whole number — no single match ever ends in 1.2 goals. That's fine: an average describes the set, not any one match. It just says the team scores a little over one goal in a typical game.

Worked example 2 — working backwards from the totals

The frequency table is a two-way street. Suppose you're told a group of pupils sat a spelling test and the mean score was 7, with \sum f = 25 pupils. What was the total of all their marks?

Rearrange \text{mean} = \dfrac{\sum f x}{\sum f} to get \sum f x = \text{mean} \times \sum f = 7 \times 25 = 175. So the whole class scored 175 marks between them. Being able to flip between "mean", "total" and "how many" is a genuinely useful skill — it's how you'd check whether one missing pupil's mark still fits.

Worked example 3 — back to the car survey

Let's finish the survey we opened with: 5 households with 1 car, 12 with 2, and 13 with 3.

Cars (x)	Households (f)	f \times x
1	5	5
2	12	24
3	13	39
Total	30	68

\sum f = 5 + 12 + 13 = 30 households (matching the survey size), and \sum f x = 5 + 24 + 39 = 68 cars in total. The mean is 68 \div 30 = 2.27 cars per household (to 2 d.p.).

Sanity check: the mean sits between the smallest value 1 and the largest 3, and leans toward 3 because most households own three cars. If your answer ever lands outside the range of the values, you've divided by the wrong thing — almost always the row-count trap below.

Two slips cause almost every wrong answer here:

Divide by \sum f, not by the number of rows. If 30 households were surveyed, you divide by 30 — the total number of data items — not by the 5 rows of the table. The rows are just the different values; the frequencies tell you how many items there really are.
Multiply first, then add. The mean is \dfrac{\sum (f \times x)}{\sum f}, so each value must be multiplied by its frequency before anything is totalled. Adding the values on their own — or adding f and x instead of multiplying them — throws the weighting away and gives nonsense.

The f \times x idea is exactly a weighted average, and once you spot it you see it everywhere:

Your course grade. If the exam is worth 70\% and coursework 30\%, your final mark weights each score by its importance — same sum, just with percentages as the "frequencies".
A cricketer's batting average and a shop's star rating are both \sum f x \div \sum f in disguise: a rating of "mostly 5 stars with a few 1s" is those star values weighted by how many people gave each.

The very same method handles grouped data too — you just use the midpoint of each class as the value x to get an estimated mean.